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The azide ions have.
(A) 20 out electrons and this isoelectronic with $B{r_2} O$
(B) 18 outer electrons and its isoelectronic with a $N{O_2} $
(C) 16 outer electrons are in isoelectronic with $C{O_2} $
(D) 14 outer electrons and is isoelectronic with ${H_2} {O_2} $

Answer
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Hint: In this question, we have been asked about the azide ion so for that, we know the formula of azide ion is ${N_3} ^ - $ and the number of electrons is 22 (one nitrogen contain 7 electrons and one due to the negative charge) so from these characteristics of azide we can solve it.

Complete step by step solution:
We have been provided with the azide ion,
Now we know azide ion has the molecular formula: ${N_3} ^ - $,
Total number of electrons present in azide ion is 22 it has 16 electrons in the outermost electron and (5 valence electrons from each nitrogen atom + 1 for a negative charge),
Total electrons are 22 (7 from each nitrogen Atom + 1 for a negative charge),
So, we can say that the total number of electrons in $C{O_2} $ are also 22 (6 from carbon and 8 from each oxygen atom),
Now $C{O_2} $ has 16 outer electrons and as $C{O_2} $ too has the same number of electrons as azide ion we can say that azide ion is isoelectronic to $C{O_2} $.

Option (C) is correct.

Note: Another way of solving the question would be comparing the properties of all the ions and the molecules given to us as in options but it will make the question a little bit longer and the chance of mistakes may increase. So, be a little bit cautious while solving it that way or try to avoid it.